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3 years ago

1. A race car moves along a circular track of radius 100m at a velocity of 25m/s. (a) What is

Physics

1 answer:

goldenfox [79]3 years ago

6 0

Given :

A race car moves along a circular track of radius 100m at a velocity of 25m/s.

To Find :

(a) What is the time taken to complete one lap of the circular track.

(b) What is the time taken for 10 laps.

Solution :

Circumference of circular track,

$C = 2\pi r\\\\C = 2\times \pi \times 100\ m\\\\C = 628.32 \ m$

a) Time taken to complete one lap is :

$t= \dfrac{628.32}{25}\ s\\\\t =25.13 \ s$

b) Time taken to complete 10 laps is :

$t_{10} = 10\times t\\\\t_{10} = 10\times 25.13\ s\\\\t_{10} = 251.3\ s$

Hence, this is the required solution.

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A floating ice block is pushed through a displacement d = (14 m) i hat - (11 m) j along a straight embankment by rushing water,

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Explanation:

Given that,

Displacement in ice block, $d=14i-11j$

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To find,

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We know that the product of force and displacement is called work done. It is also equal to the dot product of force and displacement as :

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3 years ago

A fire truck has a searchlight with a resistance of 60 (ohm) which is placed across a 24-V battery. What is the current in this

Mariulka [41]

Answer: 0.4 Amps

Explanation:

Voltage of battery = 24 Volts

Current I = ?

Resistance of searchlight (R)= 60ohms (Ω is the symbol for ohms)

Then, apply the formula for ohms law

Voltage = Current x resistance

i.e V = I x R

24V = I x 60Ω

I = 24V / 60Ω

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3 years ago

Radiant heat makes it impossible to stand close to a hot lava flow. Calculate the rate of heat loss by radiation from 1.00 m^2 o

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The rate of heat loss by radiation is equal to <u>-207.5kW</u>

Why?

To calculate the heat loss rate (or heat transfer rate) by radiation, from the given situation, we can use the following formula:

$HeatLossRate=E*S*A*((T_{cold})^{4} -(T_{hot})^{4} )$

Where,

E, is the emissivity of the body.

A, is the area of the body.

T, are the temperatures.

S, is the Stefan-Boltzmann constant, which is equal to:

$5.67x10^{-8}\frac{W}{m^{2}*K^{-4} }$

Now, before substitute the given information, we must remember that the given formula works with absolute temperatures (Kelvin), so, we need to convert the given values of temperature from Celsius degrees to Kelvin.

We know that:

$K=Celsius+273.15$

So, converting we have:

$T_{1}=1110\°C+273.15=1383.15K\\\\T_{2}=36.2\°C+273.15=309.35K$

Therefore, substituting the given information and calculating, we have:

$HeatLossRate=E*S*A*((T_{cold})^{4} -(T_{hot})^{4} )$

$HeatLossRate=1*5.67x10^{-8}\frac{W}{m^{2}*K^{-4} }*1m^{2} *((309.35K)^{4} -(1383.15})^{4} )\\\\HeatLossRate=5.67x10^{-8}\frac{W}{K^{-4} }*(95697.42K^{4} -3.66x10^{12}K^{4})\\ \\HeatLossRate=5.67x10^{-8}\frac{W}{K^{-4} }*(-3.66x10^{12} K^{4})=-207522W=-207.5kW$

Hence, we have that the rate of heat loss is equal to -207.5kW.

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3 years ago